jmc

Let $y=\frac{x^2+4x}{x-1}.$ Then we can write the given equation as $$y+\frac{72}{y}-18=0,$$so $y^2-18y+72=0.$ This factors $(y-6)(y-12)=0,$ so $y=6$ or $y=12.$

If $\frac{x^2+4x}{x-1}=6,$ then $x^2+4x=6x-6,$ or $x^2-2x+6=0.$ This quadratic has no real solutions.

If $\frac{x^2+4x}{x-1}=12,$ then $x^2+4x=12x-12,$ or $x^2-8x+12=0.$ This factors as $(x-2)(x-6)=0,$ so the solutions are $\boxed{2,6}.$